zlacker

Incidentally, while it focuses on values that fit in machine registers, you might check out Hacker's Delight's section on computing "ilog10". In the second edition, section 11-4 is on integer logarithms.

If your target language is Python3, you have the "bit_length()" method of integer objects. Multiply by c ~= 0.30103 and then correct for being too low; superficially, it seems like this algorithm should work for extremely large numbers, at least to millions of digits.

As far as test cases, n10, (n10)+1 and (n10)-1 seem like good choices and as long as the underlying mathematical function is monotonic should ferret out failures. I could easily run my implementation for numbers up to 30,000 digits, but going to 300,000 digits didn't finish while writing this message.

    log10_2 = math.log10(2)
    def digit_length_clever3(n):
        if not n: return 1
        est = math.floor(n.bit_length() * log10_2)
        low = n >= 10**est
        return est + low